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I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue-to-digital converter (ADC) with some input protection and biasing. The ADC samples at a constant/fixed 50 kSa/s.

My approach is to inject a 2 V sine wave of variable frequency \$f_{in}\$ and take the maximum (\$Vmax\$) and minimum (\$Vmin\$) samples. That gives me \$Vpp = Vmax - Vmin\$, and I sweep \$fin\$ until \$ Vpp = 2 \sqrt{\frac{1}{2}}\ V \approx 1.41\ V \$.

I've read the Wikipedia articles Bandwidth (signal processing) and Half-power point, but I remain confused.

When I find the -3 dB point, what is the corresponding official bandwidth I've found? Is it \$fin\$ or \$ \frac{fin}{2} \$ with consideration of the Nyquist rate?

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    \$\begingroup\$ It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at \$f_s / 4\$, using \$V_{pp}\$ gets problematical. \$\endgroup\$
    – TimWescott
    Commented Sep 10, 2019 at 22:42
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    \$\begingroup\$ @TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question? \$\endgroup\$
    – Paul Uszak
    Commented Sep 10, 2019 at 22:46
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    \$\begingroup\$ It would be good as a new question. The answer is semi-long. \$\endgroup\$
    – TimWescott
    Commented Sep 10, 2019 at 23:20
  • \$\begingroup\$ to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest. \$\endgroup\$
    – analogsystemsrf
    Commented Sep 11, 2019 at 4:04
  • \$\begingroup\$ Typically, an ADC circuit will have a low-pass filter in front of the ADC to minimize aliasing distortion. Ideally, this filter would block everything above the Nyquist frequency (half the sampling frequency). Of course this is not possible in real life, but one can get reasonably close to get a useful circuit. The effect of such a filter is easy to measure with an ADC. Without such a filter, you may measure how accurate your ADC measures short pulses (like analogsystemsrf explained). With a sample&hold circuit, measurement way above the sampling rate may be possible. \$\endgroup\$
    – Klaws
    Commented Sep 11, 2019 at 11:02

3 Answers 3

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The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.

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  • \$\begingroup\$ That simple...? \$\endgroup\$
    – Paul Uszak
    Commented Sep 10, 2019 at 22:20
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    \$\begingroup\$ Yes. The bandwidth does not depend on sampling rates and Nyquist rates. \$\endgroup\$
    – Barry
    Commented Sep 10, 2019 at 23:46
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    \$\begingroup\$ It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency. \$\endgroup\$
    – Caleb Reister
    Commented Sep 11, 2019 at 1:08
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The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.

enter image description here

The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way to tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.

Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.

When I find the -3dB point, what is the corresponding official bandwidth I've found? Is it fin or fin/2 with consideration of the Nyquist rate?

I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.

You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.

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  • \$\begingroup\$ Aliasing had occurred to me actually. I have considered including a small pseudo random delay into the sampling loop, say uniformly distributed between 0 and 200us. Kinda sampling scope like. \$\endgroup\$
    – Paul Uszak
    Commented Sep 11, 2019 at 22:35
  • \$\begingroup\$ So long as the frequencies aren't in phase and the measurement time is long enough you should not see any attenuation due to sampling. I simulated an ADC sampling at 50kHz with a 24.999kHz input signal. The peak and rms outputs both equaled the input over a 1 second period, but there were only 2 peaks in that time and the waveform was nothing like a sine wave! To get sine wave output it would need to be low-pass filtered, which would limit the bandwidth to < fs/2. \$\endgroup\$
    – Bruce Abbott
    Commented Sep 12, 2019 at 2:41
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I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.

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  • \$\begingroup\$ Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2? \$\endgroup\$
    – Paul Uszak
    Commented Sep 10, 2019 at 22:21
  • \$\begingroup\$ I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth. \$\endgroup\$
    – Elliot Alderson
    Commented Sep 10, 2019 at 22:54
  • \$\begingroup\$ I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal? \$\endgroup\$
    – Paul Uszak
    Commented Sep 10, 2019 at 22:59
  • \$\begingroup\$ Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC. \$\endgroup\$
    – Elliot Alderson
    Commented Sep 10, 2019 at 23:12

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